The oxide of a metal (R ) can be reduced by the metal (P) and metal (R ) can reduce the oxide of metal (Q). Then the decreasing order of the reactivity of metal (P), (Q) and (R ) with oxygen is :

Options

A. `P gt Q gt R`
B. `P gt R gt Q`
C. `R gt P gt Q`
D. `Q gt P gt R`

Answer

To solve the problem, we need to analyze the relationships between the metals P, Q, and R based on their ability to reduce each other's oxides. Here’s the step-by-step solution:

### Step 1: Understand the relationships
- The problem states that the oxide of metal R can be reduced by metal P. This means that metal P has a higher reactivity with oxygen than metal R.

### Step 2: Establish the first relationship
- From the first statement, we can conclude:
\[ \text{Reactivity of P} > \text{Reactivity of R} \]

### Step 3: Analyze the second relationship
- The problem also states that metal R can reduce the oxide of metal Q. This indicates that metal R has a higher reactivity with oxygen than metal Q.

### Step 4: Establish the second relationship
- From the second statement, we can conclude:
\[ \text{Reactivity of R} > \text{Reactivity of Q} \]

### Step 5: Combine the relationships
- Now we have two inequalities:
1. \( \text{P} > \text{R} \)
2. \( \text{R} > \text{Q} \)

- From these inequalities, we can combine them to establish the overall order of reactivity:
\[ \text{P} > \text{R} > \text{Q} \]

### Step 6: Conclusion
- Therefore, the decreasing order of reactivity of metals P, Q, and R with oxygen is:
\[ \text{P} > \text{R} > \text{Q} \]

### Final Answer
- The correct order is: **P > R > Q**

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